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Math Help - Derivatives

  1. #1
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    Derivatives

    Need derivatives for function:

    a) f(x)= ln(e^-x + 1/x)
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  2. #2
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    Use the chain rule. Here is little help to begin with:

    ln(e^{-x} + 1/x)=ln(u)'u'=(1/u)u'=\frac{1}{e^{-x} + 1/x}u'

    Now just find u'=(e^{-x} + 1/x)'=?.

    Regards.
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  3. #3
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    So it would be = \frac{1}{e^-x + \frac{1}{x}}*(e^-x(-1)-\frac{1}{x^2}) ???
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  4. #4
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    Quote Originally Posted by Stud778 View Post
    So it would be = \frac{1}{e^{-x} + \frac{1}{x}}*(-e^{-x}-\frac{1}{x^2}) ???
    Your LaTeX needs work, but that looks right to me.
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  5. #5
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    Something is wrong, because the correct answer is

    \frac{-(x-1)}{e^x+x}-\frac{1}{x}
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  6. #6
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    a
    Quote Originally Posted by shenanigans87 View Post
    Something is wrong, because the correct answer is

    \frac{-(x-1)}{e^x+x}-\frac{1}{x}
    \frac{-(x-1)}{e^x+x}-\frac{1}{x}

    \frac{-x^2+x-x-e^x}{xe^x+x^2}

    \frac{-x^2-e^x}{xe^x+x^2}

    \frac{ \frac{-x^2-e^x}{x^2e^x}}{ \frac{xe^x+x^2}{x^2e^x}}

     \frac{ - \frac{1}{e^x} - \frac{1}{x^2}}{ \frac{1}{e^x} + \frac{1}{x}}

     \frac{ -e^{-x} - \frac{1}{x^2}}{ e^{-x} + \frac{1}{x}}

    This argument follows a lot easier if you read it backwards, though.

    I kind of like that simplification, though.
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